Answer:
Complexity.
Explanation:
Okay, I am NOT explaining this whole thing, but simply, C2n product.
Don't ask. Just don't.
Answer:
The answer is c. 173 g
Explanation:
You know the reaction :
KClO3 ⇒ 2 KCl + 3 O2
By stoichiometry, that is, the amount of reagents and products in a chemical reaction when it is balanced (as in this case), it is known that for 2 moles of O2, 1 mole of KCLO3 is needed. So you can do the following rule of three to know the number of moles to produce 4.26 moles of 02:
If 1 mole of KClO3 is necessary to produce 3 moles of O2, how many moles are needed to produce 4.26 moles of 02?
So you need 1.42 moles of KClO3
Now it is necessary to know the molar mass of KClO3, which is the mass that contains 1 mole of the substance. For that you need to know the mass of K, Cl and O:
So, the molar mass of KClO3 is:
39 g/mol + 35.45 g/mol + 3*16 g/mol=122.45 g/mol
because it contains 1 atom of K, 1 atom of Cl and 3 atoms of O.
Now, to calculate the mass representing 1.42 moles of KClO3 (needed to produce 4.26 moles of O2) you simply multiply that amount of moles by the molar mass:
This means that approximately 174 g of KClO3 are necessary to produce 4.26 moles of O2.
b. nucleotide.
c. peptide.
d. monomer.
Answer:
λ = 0.0167 m = 16.7 mm
Explanation:
The wavelength of these radio waves can be found out by using the formula for the speed of radio waves:
v = fλ
where,
v = speed of radio waves = speed of light = 3 x 10⁸ m/s
f = frequency of radio waves = 18 GHz = 18 x 10⁹ Hz
λ = Wavelength = ?
Therefore,
3 x 10⁸ m/s = (18 x 10⁹ Hz)λ
λ = (3 x 10⁸ m/s)/(18 x 10⁹ Hz)
λ = 0.0167 m = 16.7 mm
Answer: The correct answer is neutrons.
Explanation:
There are 3 subatomic particles that are present in an atom. They are: protons, electrons and neutrons.
Protons carry positive charge and are found inside the nucleus of an atom.
Electrons carry negative charge and are found around the nucleus in the orbits.
Neutrons does not carry any charge and are found inside the nucleus of an atom.
Hence, the correct answer is neutrons.